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horrorfan [7]
3 years ago
14

A mixture consists of sand and an aqueous salt

Chemistry
2 answers:
Alenkinab [10]3 years ago
7 0

Answer is (4).<span>
<em>
</em>
</span>

<span><em>Explanation:</em>
</span><span>
<span>The given mixture contains an </span>insoluble solid<span> <span>and an </span></span>aqueous solution of salt.  <span>The insoluble solid is </span>sand. <span>

First </span><span>we have to separate </span>insoluble solid. <span>Sand can be separated by doing </span>filtration. When we filter the mixture sand can be seen as the residue on the filter paper. 

<span>After filtering the mixture, we should collect the </span>filtrate. <span>Filtrate is the </span>salt solution. <span>By doing </span>evaporation <span>we can get the </span>solid salt. <span>

First </span>and second choices are wrong <span>because </span>after evaporating water filtration cannot be done and salt and sand will be mixed together.<span>

Salt cannot be filtered out because the salt is soluble and it is <span>in aqueous medium. Hence, third choice is wrong</span></span></span>

Papessa [141]3 years ago
6 0
The best answer is 4  because if you get rid of the sand first and evaporate the water you're left with salt
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<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

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Solution

Graham's Law

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

r₁=4 x r₂

r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas​

Input the value :

\tt \dfrac{4r_2}{r_2}=\sqrt{\dfrac{M_2}{4} }\\\\16=\dfrac{M_2}{4}\\\\M_2=64~g/mol

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In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i
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Explanation:

Given: V_{1} = 21.35 mL,        M_{1} = 0.150 M

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Formula used to calculate molarity of H_{2}SO_{4} is as follows.

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Substitute the values into above formula as follows.

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As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = V_{1} + V_{2}

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= 0.04636 L

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